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CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

1. CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

A container contains 60 litres of milk. From this container 6 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container (in liters)?

(a)
46.34
(b)
47.36
(c)
48
(d)
43.74

Answer: Option D

Explanation :

6 liters is taken out from 60 liter container i.e. 6/60 = 1/10^th is removed. Milk removed will also be 1/10^th.

Since 1/10th is milk is removed, milk remaining after 1st replacement will be 9/10^th of 60 liters.

Similarly, amount of milk left after 3 operations = $60 {[1 - \frac{6}{60}]}^{3}$ liters

= $(60 \times \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10})$ = 43.74 liters

Hence, option (d).

2. CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

A milkman puts 120 litres of pure milk in a container. He removes 30 litres of the solution from the container and replaces it with an equal quantity of water. Thereafter, he again removes 30 litres of the solution from the container and replaces it with an equal quantity of water. What will be the concentration of milk in the container now?

If he repeats this process till the concentration of milk in the container falls below 30%, then how many times from the beginning does he carry out the process of replacing the solution in the container with water?

(a)
50%, 4 times
(b)
56.25%, 4 times
(c)
56.25%, 5 times
(d)
56.25%, 6 times

Answer: Option C

Explanation :

Initial concentration of milk = 100%

We know, Final concentration = Initial concentration × ${(1 - \frac{volume removed}{total volume})}^{n}$

As 30 litres of solution is replaced each time with an equal quantity of water concentration of the solution after the n^th replacement can be given as ${(1 - \frac{30}{120})}^{n} \times 100 = {(\frac{3}{4})}^{n} \times 100$ where n is a natural number.

So, if n = 2, ${(\frac{3}{4})}^{2} \times 100 = \frac{9}{16} \times 100$ = 56.25%

We also need to find out n if ${(\frac{3}{4})}^{n} \times 100$ < 30 or ${(\frac{3}{4})}^{n} < \frac{3}{10}$ = 0.3

For n = 3, ${(\frac{3}{4})}^{3} = \frac{27}{64}$ > 0.3

For n = 4, ${(\frac{3}{4})}^{4} = \frac{81}{256}$ > 0.3

For n = 5, ${(\frac{3}{4})}^{5} = \frac{243}{1024}$ < 0.3

So, after 5 times, concentration of milk in the solution falls below 30%.

Hence, option (c).

3. CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

From a container containing pure ghee, 30 litres are drawn out and then replaced with dalda. This process is repeated once more. The resultant mixture contains ghee and dalda in the ratio 16 : 9. Initially, what was the quantity of pure ghee in the solution (in liters)?

(a)
37.5
(b)
60
(c)
120
(d)
150

Answer: Option D

Explanation :

Ratio of ghee and dalda in final mixture is 16 : 9.
∴ Fraction of ghee in final mixture = 16/25

Let the initial quantity of pure ghee be ‘X’ litres

As per given data, 30 litres of the solution was replaced with dalda

∴ Fraction of ghee removed = $\frac{30}{X}$

while, fraction of ghee remaining = $(1 - \frac{30}{X})$

Since this process was done twice.

∴ Fraction of ghee remaining = ${(1 - \frac{30}{X})}^{2} = \frac{16}{25}$

⇒ $\frac{X - 30}{X} = \frac{4}{5}$

⇒ X = 150

Therefore, total quantity of pure ghee was 150 litres.

Hence, option (d).

4. CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

A vessel is filled with liquid, 2 parts of which are water and 3 parts oil. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half oil?

(a)
1/3
(b)
1/4
(c)
1/6
(d)
1/5

Answer: Option C

Explanation :

Suppose the vessel initially contains 5 litres of liquid.

Let x litres of this liquid be replaced with water.

Quantity of water removed will be 2/5th of x and quantity of oil removed will be 3/5th of x.

Quantity of water in new mixture = $(2 - \frac{2 x}{5} + x)$ liters

Quantity of oil in new mixture = $(3 - \frac{3 x}{5})$ liters

∴ $(2 - \frac{2 x}{5} + x) = (3 - \frac{3 x}{5})$

⇒ x = 5/6

So, part of the mixture replaced = $\frac{\frac{5}{6}}{5} = \frac{1}{6}$

Alternately,

Initial proportion of oil = 3/5

Final proportion of oil = 1/2

Let r liters is removed and replaced from V liters of liquid.

⇒ $\frac{1}{2} = \frac{3}{5} (1 - \frac{r}{V})$

⇒ fraction removed = $\frac{r}{V} = \frac{1}{6}$

Hence, option (c).

5. CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

A can contains a mixture of two liquids A and B is the ratio 5 : 3. When 8 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 5 : 7. How many litres of liquid A was contained by the can initially?

(a)
10
(b)
20
(c)
15
(d)
25

Answer: Option C

Explanation :

Suppose the can initially contains 15x and 9x of mixtures A and B respectively.

Quantity of A in mixture left = $(5 x - \frac{5}{8} \times 8)$ = 5x - 5 liters

Quantity of B in mixture left = $(3 x - \frac{3}{8} \times 8)$ = 3x - 3 liters

∴ $\frac{5 x - 5}{3 x - 3 + 8} = \frac{5}{7}$

⇒ 35x – 35 = 15x + 25

⇒ x = 3

So, the can contained 15 litres of A.

Hence, option (c).

6. CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is:

(a)
1/3
(b)
2/3
(c)
2/5
(d)
3/5

7. CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

10 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65. How much wine did the cask hold originally?

(a)
30 liters
(b)
40 liters
(c)
35 liters
(d)
45 liters

Answer: Option A

Explanation :

Let the quantity of the wine in the cask originally be x litres.

Then, quantity of wine left in cask after 4 operations = $x {[1 - \frac{10}{x}]}^{4}$ liters

∴ $[\frac{x {(1 - \frac{10}{x})}^{4}}{x}] = \frac{16}{81}$

⇒ ${(1 - \frac{10}{x})}^{4} = {(\frac{2}{3})}^{4}$

⇒ $1 - \frac{10}{x} = \frac{2}{3}$

⇒ x = 30

Hence, option (a).

8. CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

From a container containing milk, 8 liters milk was drawn out and was replaced by water. Again 8 liters of mixture was drawn out and was replaced by the water. Thus, the quantity of milk and water in the container after these two operations is 9 : 16. The quantity of mixture is:

(a)
25
(b)
26
(c)
20
(d)
31

9. CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

Nine litres are drawn from a vessel full of wine and it is then filled with water; 9 litres of the mixture are drawn and the vessel is again filled with water. If the quantity of wine now in the vessel be to the quantity of water in it as 16 : 9, how much does the vessel hold?

(a)
40 liters
(b)
50 liters
(c)
45 liters
(d)
42 liters

Answer: Option C

Explanation :

When r liters are taken out from V liters of pure wine initially and replaced with water and this process is repeated a total of n times, the quantity of wine left = $V {(1 - \frac{r}{V})}^{2}$

Let x be the total quantity of wine initially.
Since, wine and water left are in the ratio of 16 : 9, the quantity of wine left = $\frac{16}{25}$ x

⇒ $x {(1 - \frac{9}{x})}^{2} = \frac{16 x}{25}$

∴ x = 45 lit

Hence, option (c).

10. CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

From a container, full of pure alcohol, 20% is replaced by water and this process is repeated two more times. At the end of third operation, the quantity of pure alcohol reduces to

(a)
40%
(b)
50%
(c)
51.2%
(d)
58.8%

Answer: Option C

Explanation :

Let pure alcohol was 100L. So,

Water is replaced 20% in per process = 20% of 100 = 20L.

Now, we use short-cut formula for it.

Quantity of Alcohol reduces to $V {[1 - (\frac{r}{V})]}^{n}$

Here,
V = Initial quantity of alcohol.
r = Replaced water in per process.
n = No. of process repeated.

= $100 {[1 - (\frac{20}{100})]}^{3}$

= $\frac{100 \times 64}{125}$ = 51.2L.

Hence, option (c).

CRE 3 - Removal & Replacement | Arithmetic - Mixture, Alligation, Removal & Replacement

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